Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm‘s runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        assert(A!=NULL && n!=0);
        vector<int> result;
        int start = 0, end = n-1, mid;
        while(start <= end)
        {
            mid = (start+end)/2;
            if(A[mid] == target)
            {
                while(mid>=0 && A[mid]==target)
                    mid--;
                result.push_back(++mid);
                while(mid<=n-1 && A[mid]==target)
                    mid++;
                result.push_back(--mid);
                return result;
            }
            else if(A[mid] < target)
                start = mid+1;
            else
                end = mid-1;
        }
        result.push_back(-1);
        result.push_back(-1);
        return result;
    }
};